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old part of S6_5 ARCHIVE
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Part of an older draft of section 6.5 in Chapter 6 (two round wires) of Phil's transmission line notes. It uses the large-argument Bessel asymptotics (NIST 10.7.8) to show the Bessel-type coefficient fm(r) reduces to -2j(a/r)^(1/2)e^{-(a-r)/δ}e^{j(r-a)/δ}. The current density Jz(r,θ) then has the same angular distribution as the bipolar surface charge density n1, matching result (6.5.21) and relying on the charge pumping boundary condition.
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(e) The relationship between Jz(r,θ) and n(θ) as a limit of (6.5.14)
Fact: In the extreme skin effect regime,
fm(r) = -2j [ (a/r)1/2 e-(a-r)/δ ej[(r-a)/δ]
fm(a) = -2j . (6.5.22)
Proof: From NIST 10.7.8 we find that for large x argument,
Jm(x) ≈ (2/πx)1/2 cos(x-mπ/2-π/4) . NIST 10.7.8
Then for x = βr = (j-1)(r/δ) from (6.5.9) and with δ << r ≤a we keep only the larger exponential term so that
Jm(x) ≈ (2/πx)1/2 cos(x-mπ/2-π/4) ≈ (2/πx)1/2 (1/2) e+r/δ ej[(r/δ)+mπ/2+π/4]
= (1/πβr)1/2 e+r/δ ej[(r/δ)+mπ/2+π/4] . (6.5.23)
From (6.5.14),
fm(r) ≡ [ - ] . (6.5.14)
We then examine the two terms
= = (a/r)1/2 e-(a-r)/δ ej[(r-a)/δ-π/2]
= (a/r)1/2 e-(a-r)/δ ej[(r-a)/δ+π/2] . (6.5.24)
Then
fm = [ - ] = (a/r)1/2 e-(a-r)/δ ej[(r-a)/δ [e-jπ/2 - e+jπ/2]
= (a/r)1/2 e-(a-r)/δ ej[(r-a)/δ [-j - j]
= -2j [ (a/r)1/2 e-(a-r)/δ ej[(r-a)/δ] .
which then verifies (6.5.22). QED
Recalling now (6.5.14),
Jz(r,θ) = (I/πa2)(βa/4) [ f0(r) + 2 Σm=1∞ (-1)m e-m|ξ| fm(r) cos(mθ) ] (6.5.14)
we immediately obtain this skin effect limit,
Jz(r,θ) = (-2j) [(a/r)1/2 e-(a-r)/δ ej[(r-a)/δ ](I/πa2)(βa/4) *
[ 1 + 2 Σm=1∞ (-1)m e-m|ξ|cos(mθ) ] .
(6.5.25)
But the second object in square brackets also appears in (6.5.5),
n1(ξ1,θ) = (q/2π)[ 1 + 2 !Syntax Error, I (-1)m e-m|ξ| cos(mθ) ] . Bipolar (A.13) (6.5.5)
allowing us to write
Jz(r,θ) = (-2j) [(a/r)1/2 e-(a-r)/δ ej[(r-a)/δ ] (I/πa2)(βa/4) (2π/q) n1(ξ1,θ)
Jz(a,θ) = (-2j) (I/πa2)(βa/4) (2π/q) n1(ξ1,θ) . (6.5.26)
Before continuing with details, we see at once that both Jz(r,θ) and Jz(a,θ) have the same angular distribution as the surface charge density n1(ξ1,θ) . Now recall that
I = q vd (4.11.19a)
β = ej3π/4 (/δ) (6.5.9)
where vd is light speed in the transmission line dielectric. Then we find, setting r = a,
Jz(a,θ) = (-2j) (I/πa2)(βa/4) (2π/q) n1(ξ1,θ)
= (-j) (I/a2)(βa) (1/q) n1(ξ1,θ) = (-j) (vd/a2)(βa) n1(ξ1,θ)
= (-j) (vd/a)([ ej3π/4 (/δ)]a) (1/a) n1(ξ1,θ)
= e-j2π/4 ej3π/4 (vd/a) (a/δ) (1/a) n1(ξ1,θ)
= e+j2π/4 (vd/a) (a/δ) (1/a) n1(ξ1,θ)
= (1+j) (vd/a) (a/δ) (1/a) n1(ξ1,θ)
= (vd/a) (1+j) (a/δ) (1/a) n1(ξ1,θ) (6.5.27)
which agrees with the result (6.5.21) of the previous section. This seems to justify the assumptions made in the derivation of that result. Both derivations rely on the charge pumping boundary condition (D.2.24).